Question: $\dfrac{ -6i - j }{ 4 } = \dfrac{ 7i + k }{ 6 }$ Solve for $i$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ -6i - j }{ {4} } = \dfrac{ 7i + k }{ 6 }$ ${4} \cdot \dfrac{ -6i - j }{ {4} } = {4} \cdot \dfrac{ 7i + k }{ 6 }$ $-6i - j = {4} \cdot \dfrac { 7i + k }{ 6 }$ Multiply both sides by the right denominator. $-6i - j = 4 \cdot \dfrac{ 7i + k }{ {6} }$ ${6} \cdot \left( -6i - j \right) = {6} \cdot 4 \cdot \dfrac{ 7i + k }{ {6} }$ ${6} \cdot \left( -6i - j \right) = 4 \cdot \left( 7i + k \right)$ Distribute both sides ${6} \cdot \left( -6i - j \right) = {4} \cdot \left( 7i + k \right)$ $-{36}i - {6}j = {28}i + {4}k$ Combine $i$ terms on the left. $-{36i} - 6j = {28i} + 4k$ $-{64i} - 6j = 4k$ Move the $j$ term to the right. $-64i - {6j} = 4k$ $-64i = 4k + {6j}$ Isolate $i$ by dividing both sides by its coefficient. $-{64}i = 4k + 6j$ $i = \dfrac{ 4k + 6j }{ -{64} }$ All of these terms are divisible by $2$ Divide by the common factor and swap signs so the denominator isn't negative. $i = \dfrac{ -{2}k - {3}j }{ {32} }$